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More Functions and Sets Rosen 1.8. Inverse Image Let f be an invertible function from set A to set B. Let S be a subset of B. We define the inverse image.

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Presentation on theme: "More Functions and Sets Rosen 1.8. Inverse Image Let f be an invertible function from set A to set B. Let S be a subset of B. We define the inverse image."— Presentation transcript:

1 More Functions and Sets Rosen 1.8

2 Inverse Image Let f be an invertible function from set A to set B. Let S be a subset of B. We define the inverse image of S to be the subset of A containing all pre-images of all elements of S. f -1 (S) = {a  A | f(a)  S} S A f(a) a B

3 Let f be an invertible function from A to B. Let S be a subset of B. Show that f -1 (S) = f -1 (S) What do we know? f must be 1-to-1 and onto S A B b1a2b2 a1 ffff f -1 (S)

4 Proof: We must show that f -1 (S)  f -1 (S) and that f -1 (S)  f -1 (S). Let x  f -1 (S). Then x  A and f(x)  S. Since f(x)  S, x  f -1 (S). Therefore x  f -1 (S). Now let x  f -1 (S). Then x  f -1 (S) which implies that f(x)  S. Therefore f(x)  S and x  f -1 (S) Let f be an invertible function from A to B. Let S be a subset of B. Show that f -1 (S) = f -1 (S)

5 Proof: f -1 (S) = {x  A | f(x)  S}Set builder notation = {x  A | f(x)  S}Def of Complement = f -1 (S)Def of Complement Let f be an invertible function from A to B. Let S be a subset of B. Show that f -1 (S) = f -1 (S)

6 Floor and Ceiling Functions The floor function assigns to the real number x the largest integer that is less than or equal to x.  x   x  = n iff n  x < n+1, n  Z  x  = n iff x-1 < n  x, n  Z The ceiling function assigns to the real number x the smallest integer that is greater than or equal to x.  x   x  = n iff n-1 < x  n, n  Z  x  = n iff x  n < x+1, n  Z

7 Examples  0.5  = 1  0.5  = 0  -0.3  = 0  -0.3  = -1  6  = 6  6  = 6  -3.4  = -3  3.9  = 3

8 Prove that  x+m  =  x  + m when m is an integer. Proof: Assume that  x  = n, n  Z. Therefore n  x < n+1. Next we add m to each term in the inequality to get n+m  x+m < n+m+1. Therefore  x+m  = n+m =  x  + m  x  = n iff n  x < n+1, n  Z

9 Let x  R. Show that  2x  =  x  +  x+1/2  Proof: Let n  Z such that  x  = n. Therefore n  x < n+1. We will look at the two cases: x  n + 1/2 and x < n + 1/2. Case 1: x  n + 1/2 Then 2n+1  2x < 2n+2, so  2x  = 2n+1 Also n+1  x + 1/2 < n+2, so  x + 1/2  = n+1  2x  = 2n+1 = n + n+1 =  x  +  x+1/2  nn+1

10 Let x  R. Show that  2x  =  x  +  x+1/2  Case 2: x < n + 1/2 Then 2n  2x < 2n+1, so  2x  = 2n Also n  x + 1/2 < n+1, so  x + 1/2  = n  2x  = 2n = n + n =  x  +  x+1/2 

11 Characteristic Function Let S be a subset of a universal set U. The characteristic function f S of S is the function from U to {0,1}such that f S (x) = 1 if x  S and f S (x) = 0 if x  S. Example: Let U = Z and S = {2,4,6,8}. f S (4) = 1 f S (10) = 0

12 Let A and B be sets. Show that for all x, f A  B (x) = f A (x)f B (x) Proof: f A  B (x) must equal either 0 or 1. Suppose that f A  B (x) = 1. Then x must be in the intersection of A and B. Since x  A  B, then x  A and x  B. Since x  A, f A (x)=1 and since x  B f B (x) = 1. Therefore f A  B = f A (x)f B (x) = 1. If f A  B (x) = 0. Then x  A  B. Since x is not in the intersection of A and B, either x  A or x  B or x is not in either A or B. If x  A, then f A (x)=0. If x  B, then f B (x) = 0. In either case f A  B = f A (x)f B (x) = 0.

13 Let A and B be sets. Show that for all x, f A  B (x) = f A (x) + f B (x) - f A (x)f B (x) Proof: f A  B (x) must equal either 0 or 1. Suppose that f A  B (x) = 1. Then x  A or x  B or x is in both A and B. If x is in one set but not the other, then f A (x) + f B (x) - f A (x)f B (x) = 1+0+(1)(0) = 1. If x is in both A and B, then f A (x) + f B (x) - f A (x)f B (x) = 1+1 – (1)(1) = 1. If f A  B (x) = 0. Then x  A and x  B. Then f A (x) + f B (x) - f A (x)f B (x) = – (0)(0) = 0.

14 Let A and B be sets. Show that for all x, f A  B (x) = f A (x) + f B (x) - f A (x)f B (x) ABA  B f A  B (x) f A (x) + f B (x) - f A (x)f B (x) (1)(1) = (1)(0) = (0)(1) = )-(0)(0) = 0


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